0 is 0 == 0 (#evaluates to True?)

Question

This baffles me. Even without knowing the precedence order, one can check that the two possible ways to gather the expression would give False :

>>> (0 is 0) == 0
False
>>> 0 is (0 == 0)
False

But

>>> 0 is 0 == 0
True

How come?

This must be a dupe, but `0 is 0 == 0` is parsed as `(0 is 0) and (0 == 0)`, just like `a == b == c` is parsed as `a == b and b == c`. — Paul Hankin, Jul 29 '16 at 07:54
Check out this post: [Understanding Python's “is” operator](http://stackoverflow.com/questions/13650293/understanding-pythons-is-operator) — Just Ice, Jul 29 '16 at 08:11

score 20 · Answer 1 · edited May 23 '17 at 12:08

You are using comparison operator chaining. The expression is interpreted as:

(0 is 0) and (0 == 0)

From the Comparisons documentation:

Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

0 is 0 is true because Python interns small integers, an implementation detail, so you get (True) and (True) producing True.

score 7 · Accepted Answer · answered Jul 29 '16 at 07:57

When chaining comparison operators in Python, the operators aren't actually applied to the result of the other operators, but are applied to the operands individually. That is x ? y ?? z (where ? and ?? are supposed to stand in for some comparison operators) is neither equivalent to (x ? y) ?? z nor x ? (y ?? z), but rather x ? y and y ?? z.

This is particularly useful for > and co., allowing you to write things like min < x < max and have it do what you want rather than comparing a boolean to a number (which would happen in most other languages).

0 is 0 == 0 (#evaluates to True?)

2 Answers2